Let … up to 40 terms. If … is a root of…

Single correct+4 / -1Answer matches NTA’s final key

Let $\alpha = 3 + 4 + 8 + 9 + 13 + 14 + \dots$ up to 40 terms. If $(\tan\beta)^{\frac{\alpha}{1020}}$ is a root of the equation $x^2 + x - 2 = 0$ , $\beta \in \left(0, \frac{\pi}{2}\right)$ , then $\sin^2\beta + 3\cos^2\beta$ is equal to:

$2$
$\frac{7}{4}$
$\frac{5}{2}$
$\frac{3}{2}$

Solution

Sum the series in pairs. $(3 + 4) + (8 + 9) + (13 + 14) + \dots$ is 20 pair-sums $7, 17, 27, \dots$ , an AP with difference 10:

$\alpha = \frac{20}{2}\big(2 \cdot 7 + 19 \cdot 10\big) = 10 \cdot 204 = 2040, \qquad \frac{\alpha}{1020} = 2.$

Use the root. $x^2 + x - 2 = (x + 2)(x - 1)$ , so the roots are $1$ and $-2$ . Since $\tan^2\beta \ge 0$ , $\tan^2\beta = 1$ , and with $\beta$ acute, $\beta = \frac{\pi}{4}$ .

$\sin^2\beta + 3\cos^2\beta = \tfrac12 + \tfrac32 = 2.$

Answer: 2.

Answer checked against the official answer key. The solution is our own working.

JEE MAIN 2026 · Sequence and Series

Solution

Keep going